[proofplan]
We use the continued fraction expansion of $\sqrt{D}$. Since $D$ is a positive nonsquare integer, $\sqrt{D}$ is irrational, so its simple continued fraction is infinite and eventually periodic. The periodic continued fraction theory of quadratic irrationals supplies a convergent $p_n/q_n$ whose numerator and denominator satisfy the Pell identity $p_n^2 - Dq_n^2 = 1$. Since convergent denominators are positive and $\sqrt{D}>0$, this gives the required positive integers.
[/proofplan]
[step:Show that $\sqrt{D}$ has an infinite periodic continued fraction]
Let
\begin{align*}
\alpha := \sqrt{D} \in \mathbb{R}_{+}.
\end{align*}
Because $D \in \mathbb{N}$ is not a square, $\alpha \notin \mathbb{Q}$. Hence the simple continued fraction expansion of $\alpha$ does not terminate. By [Lagrange's theorem](/theorems/841) for quadratic irrationals, $\alpha$ has an eventually periodic simple continued fraction expansion. Since $\alpha = \sqrt{D}$ is a reduced quadratic irrational after the initial integral part, its continued fraction has the form
\begin{align*}
\sqrt{D} = [a_0; \overline{a_1, a_2, \dots, a_\ell}]
\end{align*}
for some $a_0 \in \mathbb{N} \cup \{0\}$, some period length $\ell \in \mathbb{N}$, and positive integers $a_1,\dots,a_\ell \in \mathbb{N}$.
(citing a result not yet in the wiki: [Lagrange's theorem](/theorems/782) on periodic continued fractions of quadratic irrationals)
[guided]
Define
\begin{align*}
\alpha := \sqrt{D} \in \mathbb{R}_{+}.
\end{align*}
The nonsquare hypothesis is used first here. If $\sqrt{D}$ were rational, then $D$ would be a square integer; since $D$ is not a square, $\sqrt{D}$ is irrational. Therefore its simple continued fraction expansion cannot terminate, because finite simple continued fractions represent rational numbers.
The structural theorem for quadratic irrationals is Lagrange's theorem: a real irrational number has an eventually periodic simple continued fraction expansion exactly when it is a quadratic irrational. The number $\sqrt{D}$ is a quadratic irrational, since it satisfies the polynomial equation $t^2-D=0$ and is irrational. Applying that theorem gives eventual periodicity. For square roots of positive nonsquare integers, the usual continued fraction reduction gives a purely periodic tail after the initial integer part, so there exist $a_0 \in \mathbb{N}\cup\{0\}$, a period length $\ell \in \mathbb{N}$, and positive integers $a_1,\dots,a_\ell \in \mathbb{N}$ such that
\begin{align*}
\sqrt{D} = [a_0; \overline{a_1, a_2, \dots, a_\ell}].
\end{align*}
(citing a result not yet in the wiki: Lagrange's theorem on periodic continued fractions of quadratic irrationals)
[/guided]
[/step]
[step:Choose the Pell-producing convergent]
For each integer $n \geq 0$, let $p_n/q_n$ denote the $n$th convergent of the simple continued fraction of $\sqrt{D}$, with $p_n \in \mathbb{Z}$ and $q_n \in \mathbb{N}$. The standard continued fraction theorem for Pell equations states that, for the periodic expansion
\begin{align*}
\sqrt{D} = [a_0; \overline{a_1,\dots,a_\ell}],
\end{align*}
there exists an index $n \geq 0$ such that
\begin{align*}
p_n^2 - Dq_n^2 = 1.
\end{align*}
More explicitly, one may take $n=\ell-1$ when $\ell$ is even, and $n=2\ell-1$ when $\ell$ is odd.
(citing a result not yet in the wiki: continued fraction criterion for solutions of Pell's equation)
[guided]
For every integer $n \geq 0$, define $p_n/q_n$ to be the $n$th convergent of the simple continued fraction expansion of $\sqrt{D}$. By construction of simple continued fractions, the numerator $p_n$ is an integer and the denominator $q_n$ is a positive integer:
\begin{align*}
p_n \in \mathbb{Z}, \qquad q_n \in \mathbb{N}.
\end{align*}
Now we use the Pell-specific consequence of periodic continued fractions. It says that if
\begin{align*}
\sqrt{D} = [a_0; \overline{a_1,\dots,a_\ell}]
\end{align*}
has period length $\ell$, then one of the convergents satisfies the exact norm equation
\begin{align*}
p_n^2 - Dq_n^2 = 1.
\end{align*}
The relevant index is determined by the parity of the period:
\begin{align*}
n =
\begin{cases}
\ell - 1, & \text{if } \ell \text{ is even},\\
2\ell - 1, & \text{if } \ell \text{ is odd}.
\end{cases}
\end{align*}
Thus there is at least one convergent whose numerator and denominator solve Pell's equation.
(citing a result not yet in the wiki: continued fraction criterion for solutions of Pell's equation)
[/guided]
[/step]
[step:Verify that the convergent gives positive integer coordinates]
Choose the index $n$ from the preceding step and define
\begin{align*}
x &:= p_n, &
y &:= q_n.
\end{align*}
Since $q_n \in \mathbb{N}$, we have $y \in \mathbb{N}$. Also $p_n/q_n$ is a convergent to the positive real number $\sqrt{D}$, and all partial quotients in the simple continued fraction of $\sqrt{D}$ are nonnegative with positive tail entries, so $p_n>0$. Hence $x \in \mathbb{N}$. The identity from the preceding step gives
\begin{align*}
x^2 - Dy^2 = p_n^2 - Dq_n^2 = 1.
\end{align*}
Therefore there exist $x,y \in \mathbb{N}$ satisfying Pell's equation, as required.
[/step]
